求通过两点且平行于给定轴的平面方程
给定两个点 A(x1，y1，z1) 和 B(x2，y2，z2) 和一组代表轴 (ai + bj + ck) 的点(A，B，c)，任务是求通过给定点 A 和 B 并与给定轴平行的平面方程。
示例:
输入: x1 = 1，y1 = 2，z1 = 3，x2 = 3，y2 = 4，z2 = 5，a= 6，b = 7，c = 8 输出: 2x + 4y + 2z + 0 = 0
输入: x1 = 2，y1 = 3，z1 = 5，x2 = 6，y2 = 7，z2 = 8，a= 11，b = 23，c = 10。 输出: -29x + 7y + 48z + 0= 0
方法: 从平面 A 和 B 上给定的两个点，方向比直线的向量方程 AB 由下式给出:
方向比=(x2–x1，y2–y1，z2–Z1)
自线
平行于给定的轴
。因此，交叉产品
和
是 0，由下式给出:
其中， d、e、f 为 AB 线向量方程的系数，即 d =(x2–x1) e =(y2–y1) f =(z2–Z1) a、b、c 为给定轴的系数。
由上述行列式形成的方程由下式给出:
(等式 1)
方程 1 与直线 AB 垂直，即与所需平面垂直。
让平面的方程式由
(方程式 2)
给出，其中 A、B、C 是平面垂直于平面的方向比。
由于方程 1 是方程 2 相互垂直，因此方程 1 & 2 的方向比数值是平行的。那么平面的系数由下式给出:
a =(B * f–C * e)、 B =(a * f–C * d)和 C =(a * e–B * d)
现在平面和向量线 AB 的点积给出 D 的值为
d =-(A * d–B * e+C * f)
下面是上述方法的实现:
C++
// C++ implementation to find the
// equation of plane which passes
// through two points and parallel
// to a given axis
#include <bits/stdc++.h>
using namespace std;
void findEquation(int x1, int y1, int z1,
int x2, int y2, int z2,
int d, int e, int f)
{
// Find direction vector
// of points (x1, y1, z1)
// and (x2, y2, z2)
double a = x2 - x1;
double b = y2 - y1;
double c = z2 - z1;
// Values that are calculated
// and simplified from the
// cross product
int A = (b * f - c * e);
int B = (a * f - c * d);
int C = (a * e - b * d);
int D = -(A * d - B * e + C * f);
// Print the equation of plane
cout << A << "x + " << B << "y + "
<< C << "z + " << D << "= 0";
}
// Driver Code
int main()
{
// Point A
int x1 = 2, y1 = 3, z1 = 5;
// Point B
int x2 = 6, y2 = 7, z2 = 8;
// Given axis
int a = 11, b = 23, c = 10;
// Function Call
findEquation(x1, y1, z1,
x2, y2, z2,
a, b, c);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to find the
// equation of plane which passes
// through two points and parallel
// to a given axis
import java.util.*;
class GFG{
static void findEquation(int x1, int y1, int z1,
int x2, int y2, int z2,
int d, int e, int f)
{
// Find direction vector
// of points (x1, y1, z1)
// and (x2, y2, z2)
double a = x2 - x1;
double b = y2 - y1;
double c = z2 - z1;
// Values that are calculated
// and simplified from the
// cross product
int A = (int)(b * f - c * e);
int B = (int)(a * f - c * d);
int C = (int)(a * e - b * d);
int D = -(int)(A * d - B * e + C * f);
// Print the equation of plane
System.out.println(A + "x + " + B + "y + " +
C + "z + " + D + "= 0 ");
}
// Driver code
public static void main(String[] args)
{
// Point A
int x1 = 2, y1 = 3, z1 = 5;
// Point B
int x2 = 6, y2 = 7, z2 = 8;
// Given axis
int a = 11, b = 23, c = 10;
// Function Call
findEquation(x1, y1, z1,
x2, y2, z2,
a, b, c);
}
}
// This code is contributed by Pratima Pandey
Python 3
# Python3 implementation
# to find the equation
# of plane which passes
# through two points and
# parallel to a given axis
def findEquation(x1, y1, z1,
x2, y2, z2,
d, e, f):
# Find direction vector
# of points (x1, y1, z1)
# and (x2, y2, z2)
a = x2 - x1
b = y2 - y1
c = z2 - z1
# Values that are calculated
# and simplified from the
# cross product
A = (b * f - c * e)
B = (a * f - c * d)
C = (a * e - b * d)
D = -(A * d - B *
e + C * f)
# Print the equation of plane
print (A, "x + ", B, "y + ",
C, "z + ", D, "= 0")
# Driver Code
if __name__ == "__main__":
# Point A
x1 = 2
y1 = 3
z1 = 5;
# Point B
x2 = 6
y2 = 7
z2 = 8
# Given axis
a = 11
b = 23
c = 10
# Function Call
findEquation(x1, y1, z1,
x2, y2, z2,
a, b, c)
# This code is contributed by Chitranayal
C
// C# implementation to find the
// equation of plane which passes
// through two points and parallel
// to a given axis
using System;
class GFG{
static void findEquation(int x1, int y1, int z1,
int x2, int y2, int z2,
int d, int e, int f)
{
// Find direction vector
// of points (x1, y1, z1)
// and (x2, y2, z2)
double a = x2 - x1;
double b = y2 - y1;
double c = z2 - z1;
// Values that are calculated
// and simplified from the
// cross product
int A = (int)(b * f - c * e);
int B = (int)(a * f - c * d);
int C = (int)(a * e - b * d);
int D = -(int)(A * d - B * e + C * f);
// Print the equation of plane
Console.Write(A + "x + " + B + "y + " +
C + "z + " + D + "= 0 ");
}
// Driver code
public static void Main()
{
// Point A
int x1 = 2, y1 = 3, z1 = 5;
// Point B
int x2 = 6, y2 = 7, z2 = 8;
// Given axis
int a = 11, b = 23, c = 10;
// Function Call
findEquation(x1, y1, z1,
x2, y2, z2,
a, b, c);
}
}
// This code is contributed by Code_Mech
java 描述语言
<script>
// javascript implementation to find the
// equation of plane which passes
// through two points and parallel
// to a given axis
function findEquation(x1 , y1 , z1 , x2 , y2 , z2 , d , e , f)
{
// Find direction vector
// of points (x1, y1, z1)
// and (x2, y2, z2)
var a = x2 - x1;
var b = y2 - y1;
var c = z2 - z1;
// Values that are calculated
// and simplified from the
// cross product
var A = parseInt( (b * f - c * e));
var B = parseInt( (a * f - c * d));
var C = parseInt( (a * e - b * d));
var D = -parseInt( (A * d - B * e + C * f));
// Print the equation of plane
document.write(A + "x + " + B + "y + " + C + "z + " + D + "= 0 ");
}
// Driver code
// Point A
var x1 = 2, y1 = 3, z1 = 5;
// Point B
var x2 = 6, y2 = 7, z2 = 8;
// Given axis
var a = 11, b = 23, c = 10;
// Function Call
findEquation(x1, y1, z1, x2, y2, z2, a, b, c);
// This code is contributed by Rajput-Ji
</script>
Output:
-29x + 7y + 48z + 0= 0
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